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Exercise 7.1 · Q12

Q.For the parabola 3y2=16x3y^2 = 16x, find the parameter of the point (27,−12)(27,-12).

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As above, a=43a=\dfrac{4}{3}. Using y=2aty=2at: −12=83t⇒t=−12×38=−92-12=\dfrac{8}{3}t \Rightarrow t=-12\times\dfrac{3}{8}=-\dfrac{9}{2}. …

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